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On consecutive primitive elements in a finite field

On consecutive primitive elements in a finite field Abstract For $$q$$ an odd prime power with $$q>169,$$ we prove that there are always three consecutive primitive elements in the finite field $$\mathbb {F}_{q}$$. Indeed, there are precisely eleven values of $$q \leq 169$$ for which this is false. For $$4\leq n \leq 8,$$ we present conjectures on the size of $$q_{0}(n)$$ such that $$q>q_{0}(n)$$ guarantees the existence of $$n$$ consecutive primitive elements in $$\mathbb {F}_{q}$$, provided that $$\mathbb {F}_{q}$$ has characteristic at least $$n$$. Finally, we improve the upper bound on $$q_{0}(n)$$ for all $$n\geq 3$$. 1. Introduction Let $$q$$ be a prime power and consider primitive elements in $$\mathbb {F}_{q}$$, the finite field of order $$q$$. Cohen [2–4] proved that $$\mathbb {F}_{q}$$ contains two consecutive distinct primitive elements whenever $$q>7$$. For $$n\geq 2,$$ we wish to determine $$q_{0}(n)$$ such that $$\mathbb {F}_{q}$$, assumed to have characteristic larger than or equal to $$n$$, contains $$n$$ consecutive distinct primitive elements for all $$q>q_{0}(n)$$. Carlitz [1] showed that $$q_{0}(n)$$ exists for all $$n$$. Tanti and Thangadurai [9, Theorem 1.3] showed that   \[q_{0}(n) \leq \exp (2^{5.54 n}), \quad (n\geq 2).\] (1.1) When $$n=3$$ this gives the enormous bound $$10^{43\,743}$$. The main point of this article is to apply techniques from [5] to prove the following result. Theorem 1 The finite field$$\mathbb {F}_{q}$$contains three consecutive primitive elements for all odd$$q>169$$. Indeed, the only fields$$\mathbb {F}_{q}$$$$($$with$$q$$odd$$)$$that do not contain three consecutive primitive elements are those for which$$q=3,5,7,9,13,25,29,61,81,121$$, or$$169$$.When $$n\geq 4$$ we improve the estimate for $$q_{0}(n)$$ in the following theorem. Theorem 2 The field$$\mathbb {F}_{q},$$assumed to have characteristic at least$$n,$$contains$$n$$consecutive primitive elements provided that$$q>q_{0}(n),$$where values of$$q_{0}(n)$$are given in the third column of Table1for$$4\leq n \leq 10$$and$$q_{0}(n)=\exp (2^{2.77 n})$$for$$n\geq 11$$. Table 1. Range of potential $$q$$ given a value of $$n$$. $$n$$  Bound on $$\omega (q-1)$$  Bound on $$q$$ ($$q_0(n)$$)  $$3$$  $$13$$  $$3.49\times 10^{15\phantom {12}}$$  $$4$$  $$23$$  $$3.29\times 10^{32\phantom {12}}$$  $$5$$  $$37$$  $$4.22\times 10^{61\phantom {12}}$$  $$6$$  $$59$$  $$4.61\times 10^{113\phantom {1}}$$  $$7$$  $$100$$  $$3.75\times 10^{220\phantom {1}}$$  $$8$$  $$171$$  $$2.27\times 10^{425\phantom {1}}$$  $$9$$  $$301$$  $$1.01\times 10^{836\phantom {1}}$$  $$10$$  $$533$$  $$6.69\times 10^{1638}$$  $$n$$  Bound on $$\omega (q-1)$$  Bound on $$q$$ ($$q_0(n)$$)  $$3$$  $$13$$  $$3.49\times 10^{15\phantom {12}}$$  $$4$$  $$23$$  $$3.29\times 10^{32\phantom {12}}$$  $$5$$  $$37$$  $$4.22\times 10^{61\phantom {12}}$$  $$6$$  $$59$$  $$4.61\times 10^{113\phantom {1}}$$  $$7$$  $$100$$  $$3.75\times 10^{220\phantom {1}}$$  $$8$$  $$171$$  $$2.27\times 10^{425\phantom {1}}$$  $$9$$  $$301$$  $$1.01\times 10^{836\phantom {1}}$$  $$10$$  $$533$$  $$6.69\times 10^{1638}$$  View Large We prove Theorem 2 and discuss the construction of Table 1 in Section 4. We remark that the outer exponent in the bound in Theorem 2 is half of that given in (1.1) owing entirely to the superior sieving inequality used in Theorem 3. The double exponent still gives an enormous bound on $$q_{0}(n)$$. Were one interested in bounds for specific values of $$n\geq 11,$$ one should extend Table 1 as per Section 4. In Section 5, we present an algorithm that, along with Theorem 2, proves Theorem 1. Whereas we are not able to resolve completely the values of $$q_{0}(n)$$ for $$n\geq 4$$, we present, in Section 6, some conjectures as to the size of $$q_{0}(n)$$ for $$4\leq n \leq 8$$. 2. Character sum expressions and estimates Let $$\omega (m)$$ denote the number of distinct prime factors of $$m$$ so that $$W(m)=2^{\omega (m)}$$ is the number of square-free divisors of $$m$$. Also, let $$\theta (m)=\prod _{p|m}(1-p^{-1})$$. For any integer $$m$$ define its radical $$\mathrm {Rad}(m)$$ as the product of all distinct prime factors of $$m$$. Let $$e$$ be a divisor of $$q-1$$. Call $$g \in \mathbb {F}_{q}$$$$e$$-free if $$g \neq 0$$ and $$g = h^d$$, where $$h \in \mathbb {F}_{q}$$ and $$d\,{|}\,e$$ implies $$d=1$$. The notion of $$e$$-free depends (among divisors of $$q-1$$) only on $$\mathrm {Rad}(e)$$. Moreover, in this terminology a primitive element of $$\mathbb {F}_{q}$$ is a $$(q-1)$$-free element. The definition of any multiplicative character $$\chi$$ on $$\mathbb {F}_{q}^* $$ is extended to the whole of $$\mathbb {F}_{q}$$ by setting $$\chi (0)=0$$. In fact, for any divisor $$d$$ of $$\phi (q-1)$$, there are precisely $$\phi (d)$$ characters of order (precisely) $$d$$, a typical such character being denoted by $$\chi _d$$. In particular, $$\chi _1$$, the principal character, takes the value $$1$$ at all non-zero elements of $$\mathbb {F}_{q}$$ (whereas $$\chi _1(0)=0$$). A convenient shorthand notation to be employed for any divisor $$e$$ of $$q-1$$ is   \[\int _{d\,{|}\,e} = \sum _{d\,{|}\,e}\frac {\mu (d)}{\phi (d)} \sum _{\chi _{d}},\] (2.1) where the sum over $$\chi _{d}$$ is the sum over all $$\phi (d)$$ multiplicative characters $$\chi _d$$ of $$\mathbb {F}_{q}$$ of exact order $$d$$, and where $$\mu (n)$$ is the Möbius function. Its significance is that, for any $$g \in \mathbb {F}_{q}$$,   \[\theta (e)\int _{d\,{|}\,e}\chi _d(g) = \left \{\begin {array}{@{}cl} 1 & \mbox {if $g$ is non-zero and $e$-free}, \\ 0 & \mbox {otherwise.} \end {array} \right .\] In this expression (and throughout) only characters $$\chi _d$$ with $$d$$ square-free contribute (even if $$e$$ is not square-free). The present investigation concerns the question of the existence of $$n \geq 2$$ consecutive distinct primitive elements in $$\mathbb {F}_{q}$$, that is, whether there exists $$g \in \mathbb {F}_{q}$$ such that $$\{g, g+1,\ldots , g+n-1\}$$ is a set of $$n$$ distinct primitive elements of $$\mathbb {F}_{q}$$. Define $$p$$ to be the characteristic of $$\mathbb {F}_q$$, so that $$q$$ is a power of $$p$$. Then, necessarily, $$n \leq p$$ and, by [5, Theorem 1], we can suppose $$n \geq 3$$. Assume therefore throughout that $$3 \leq n\leq p$$. In particular, $$q$$ is odd. Let $$e_1, \ldots , e_n$$ be divisors of $$q-1$$. (In practice all divisors will be even.) Define $$N(e_1, \ldots , e_n)$$ to be the number of (non-zero) $$g \in \mathbb {F}_{q}$$ such that $$g+k-1$$ is $$e_k$$-free for each $$k=1, \ldots , n$$. The first step is the standard expression for this quantity in terms of the multiplicative characters of $$\mathbb {F}_{q}^* $$. Lemma 1 Suppose$$3 \leq n\leq p$$and$$e_1, \ldots , e_n$$are divisors of$$q-1$$. Then  \[N(e_1, \ldots , e_n)= \theta (e_1)\cdots \theta (e_n) \int _{d_1|\,e_1}\cdots \int _{d_n|\,e_n}S(\chi _{d_1}, \ldots ,\chi _{d_n}),\] where  \[S(\chi _{d_1}, \ldots ,\chi _{d_n})= \sum _{g \in \mathbb {F}_{q}}\chi _{d_1}(g) \cdots \chi _{d_n}(g+n-1).\]We now provide a bound on the size of $$S(\chi _{d_1}, \ldots ,\chi _{d_n})$$. Lemma 2 Suppose$$d_1, \ldots , d_n$$are square-free divisors of$$q-1$$. Then  \[S(\chi _{d_1}, \ldots ,\chi _{d_n})=q-n \quad \mbox {if $d_1= \cdots = d_n=1$,}\]and otherwise  \[|S(\chi _{d_1}, \ldots ,\chi _{d_n})| \leq (n-1)\sqrt {q}.\] Proof We can assume not all of $$d_1, \ldots , d_n$$ have the value $$1$$. Then $$d:= \mathrm {lcm}(d_1, \ldots , d_n)$$ is also a square-free divisor of $$q-1$$ and $$d>1$$. Evidently, there are positive integers $$c_1,\ldots ,c_n$$ with $$\gcd (c_k,d_k)=1$$ such that   \[S(\chi _{d_1}, \ldots ,\chi _{d_n})= \sum _{g \in \mathbb {F}_{q}}\chi _{d}(f(g)),\] where $$f(x) =x^{c_1}(x+1)^{c_2} \cdots (x+n-1)^{c_n}$$. Since the radical of the polynomial $$f$$ has degree $$n$$ the result holds by Weil's theorem (see [6, Theorem 5.41, p. 225] and also [7]). When $$e_1= e_2=\cdots =e_n=e$$, say, we shall abbreviate $$N(e, \ldots ,e)$$ to $$N_n(e)$$. We obtain a lower bound for $$N_{n}(e)$$ in the following lemma. Lemma 3 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Then  \[N_n(e) \geq \theta (e)^n (q-(n-1)W(e)^n\sqrt {q}).\] Proof The presence of the Möbius function in the integral notation in (2.1) means that we need only concern ourselves with the square-free divisors $$(d_{1}, \ldots , d_{n})$$ of $$e$$. The contribution of each $$n$$-tuple is given by Lemma 2. Hence, $$N_n(e) \geq \theta (e)^n (q-n-(n-1)(W(e)^n-1)\sqrt {q})$$. The result will follow provided that $$\sqrt {q}(n-1) - n$$ is positive. This is easy to see, since $$n\geq 3$$ and $$q \geq 3$$. Applying Lemma 3 with $$e=q-1$$ gives the basic criterion that guarantees $$n$$ consecutive primitive elements for sufficiently large $$q$$. Theorem 3 Suppose$$3 \leq n\leq p$$. Suppose  \[q \geq (n-1)^2\,W(q-1)^{2n}=(n-1)^2\,2^{2n\omega (q-1)}.\] (2.2)Then there exists a set of$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. As an application of Theorem 3, consider the case $$n=3$$. Let $$P_m$$ be the product of the first $$m$$ primes. A quick numerical computation reveals that for $$m\geq 50$$ one has $$P_m+1 \geq 2^{2+6m}$$. Hence, it follows that $$\mathbb {F}_q$$ has three consecutive primitive elements when $$\omega (q-1)\geq 50$$ or when $$q\geq 2^{2+6\times 50}$$. We shall soon improve this markedly. We now briefly present an improvement in the above discussion when $$q \equiv 3\pmod 4$$. The improvement in this case, in which $$-1$$ is a non-square in $$\mathbb {F}_{q}$$, is related to a device used in [4]. Note that, in Lemma 1, in the definition of $$S(\chi _{d_1}, \ldots ,\chi _{d_n})$$ we may replace $$g$$ by $$-g-(n-1)$$ and obtain   \[\begin {array}{rl} S(\chi _{d_1}, \ldots ,\chi _{d_n}) & = \sum _{g \in \mathbb {F}_{q}} \prod _{i=1}^n \chi _{d_i}(-g-n+1+i-1)= \sum _{g \in \mathbb {F}_{q}} \prod _{i=1}^n \chi _{d_i}(-1)\chi _{d_i}(g+n-i) \\ & = (\chi _{d_1}\chi _{d_2}\cdots \chi _{d_n})(-1) S(\chi _{d_n}, \ldots ,\chi _{d_1}). \end {array}\] In particular, if an odd number of the integers $$d_1, \ldots , d_n$$ are even, then $$S(\chi _{d_n}, \ldots ,\chi _{d_1})=-S(\chi _{d_1}, \ldots ,\chi _{d_n})$$. As a consequence, in Lemma 1, if $$e_1= \cdots =e_n=e$$ is even, then, on the right-hand side of the expression for $$N_n(e)$$, the terms corresponding to divisors $$(d_1, \ldots , d_n)$$ with an odd number of even divisors cancel out exactly and only those with an even number of even divisors contribute. This accounts for precisely half the $$n$$-tuples $$(d_1, \ldots , d_n)$$. Accordingly, we obtain the following improvement of Lemma 3 and Theorem 3 in this situation. Theorem 4 Suppose$$3\leq n\leq p$$and$$e$$is an even divisor of$$q-1,$$where$$q\equiv 3 \pmod 4$$and$$q \geq 7$$. Then  \[N_n(e) \geq \theta (e)^n \left ( q- \frac {n-1}{2}\,W(e)^n\sqrt {q}\right ) .\]Moreover, if  \[q \geq \frac {(n-1)^2}{4}W(q-1)^{2n}= (n-1)^2 2^{2(n\omega (q-1)-1)},\]then there exists a set of$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. 3. Sieving inequalities and estimates As before, assume $$3 \leq n \leq p$$ and let $$e$$ be a divisor of $$q-1$$. Given positive integers $$m, j, k$$ with $$1 \leq j,k \leq n$$ define $$m_{jk}=m$$ if $$j=k$$ and otherwise $$m_{jk}=1$$. Lemma 4 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Let$$l$$be a prime divisor of$$q-1$$not dividing$$e$$. Then, for each$$j \in \{1, \ldots , n\},$$  \[|N(l_{j_1}e, \ldots l_{jn}e) -\theta (l)N_n(e)| \leq (1-1/l)\,\theta (e)^{n}(n-1)W(e)^n \sqrt {q}.\] Proof Observe that $$\theta (le)= (1-1/l)\,\theta (e)$$ and that, by Lemma 1, all the character sums in $$\theta (l)N_n(e)$$ appear identically in $$N(l_{j_1}e, \ldots l_{jn}e)$$. Hence, by Lemma 2,   \[|N(l_{j_1}e, \ldots l_{jn}e) -\theta (l)N_n(e)| \leq (1-1/l)\,\theta (e)^{n}(n-1)W(e)^{n-1} (W(le)-W(e)) \sqrt {q}\] and the result follows since $$W(le)=2W(e)$$. This leads to the main sieving result. Let $$e$$ be a divisor of $$q-1$$. In what follows, if $$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$ set $$s=0$$ and $$\delta =1$$. Otherwise, let $$p_1, \ldots , p_s$$, $$s \geq 1$$, be the primes dividing $$q-1$$ but not $$e$$, and set $$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. It is essential to choose $$e$$ so that $$\delta >0$$. Lemma 5 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Then, with the above notation,  \[N_n(q-1) \geq \left ( \sum _{j=1}^n\sum _{i=1}^sN(p_{ij1}e, \ldots , p_{ijn}e) \right ) - (ns-1)N_n(e),\] (3.1)where$$p_{ijk}$$means$$(p_i)_{jk}$$. Hence  \[N_n(q-1) \geq \delta N_n(e) + \sum _{i=1}^s \left ( \sum _{j=1}^n N(p_{ij1}e, \ldots , p_{ijn}e) -\theta (p_i)N_n(e) \right ) .\] (3.2) Proof It suffices to suppose $$s \geq 1$$. The various $$N$$ terms on the right-hand side of (3.1) can be regarded as counting functions on the set of $$g \in \mathbb {F}_{q}$$ for which $$g+j-1$$ is $$e$$-free for each $$j=1, \ldots , n$$. In particular, $$N_n(e)$$ counts all such elements, whereas, for each $$i=1, \ldots , s$$, and $$1 \leq j \leq n$$, $$N(p_{ij1}e,\ldots , p_{ijn}e)$$ counts only those for which additionally $$g+j-1$$ is $$p_i$$-free. Note that $$N_n(q-1)$$ is the number of elements $$g$$ such that not only are $$g+j-1$$$$e$$-free for each $$j=1, \ldots , n$$ but additionally $$g+j-1$$ is $$p_i$$-free for each $$1\leq i\leq s,\ 1 \leq j \leq n$$. Hence we see that, for a given $$g \in \mathbb {F}_{q}$$, the right-hand side of (3.1) clocks up $$1$$ if $$g+k-1$$ is primitive for every $$1 \leq k\leq n$$, and otherwise contributes a non-positive (integral) quantity. This establishes (3.1). Since $$\theta (p_i)=1-1/p_i$$, the bound (3.2) is deduced simply by rearranging the right-hand side of (3.1). We are now able to provide a condition that, if satisfied, is sufficient to prove the existence of $$n$$ consecutive primitive elements. Theorem 5 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. If$$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$, then set$$s=0$$and$$\delta =1$$. Otherwise, let$$p_1, \ldots , p_s,$$$$s \geq 1,$$be the primes dividing$$q-1$$but not$$e$$and set$$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. Assume$$\delta >0$$. If also  \[q >\left ( (n-1)\left ( \frac {ns-1}{\delta }+2\right ) W(e)^n\right ) ^2,\] (3.3)then there exist$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. Proof Assume $$\delta >0$$. From (3.2) and Lemmas 3 and 4 (noting that for each $$j=1,\ldots , n$$ the contribution to (3.2) is the same),   \[\begin {array}{l} N_n(q-1) \geq \theta (e)^n\left ( \delta (q-(n-1)W(e)^n\sqrt {q})- n\smash {\sum _{i=1}^s} \left ( 1-\frac {1}{p_i}\right ) (n-1)W(e)^n\sqrt {q}\right ) \\ \quad \quad \quad \quad \quad = \delta \,\theta (e)^n\sqrt {q} \left ( \sqrt {q}-(n-1)W(e)^n-(n-1) \left ( \frac {ns-1}{\delta }+1\right ) W(e)^n\right ) . \end {array}\] The conclusion follows. We conclude this section with the slight improvement when $$q \equiv 3 \pmod 4$$. Now, when $$e$$ is even, the character sum expressions for the terms $$\sum _{j=1}^n N(p_{ij1}e, \ldots , p_{ijn}e) -\theta (p_i)N_n(e)$$ in (3.2) cancel unless an even number of divisors $$d_1, \ldots , d_n$$ are even. This reduces the bound in Lemma 4 by a factor of two, and gives the following improvement to Theorem 5. Theorem 6 Suppose$$q \equiv 3 \pmod 4,$$$$q \geq 7,$$$$3 \leq n\leq p$$and$$e$$is an even divisor of$$q-1$$. If$$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$set$$s=0$$and$$\delta =1$$. Otherwise, let$$p_1, \ldots , p_s,$$$$s \geq 1,$$be the primes dividing$$q-1$$but not$$e$$and set$$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. Assume$$\delta >0$$. If also  \[q >\left ( \frac {n-1}{2}\left ( \frac {ns-1}{\delta }+2\right ) W(e)^n\right ) ^2,\]then there exist$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. Remark 1 In Theorems 5 and 6, for a given $$s$$ the best (largest) value of $$\delta$$ is obtained when the largest $$s$$ prime factors of $$q-1$$ are used to compute $$\delta$$. 4. Application of Theorems 3 and 5 for generic $$n;$$ proof of Theorem 2 As an application of Theorem 5, consider the case $$n=3$$. We showed after Theorem 3 that $$\mathbb {F}_{q}$$ contains three consecutive primitive elements for all $$q$$ satisfying $$\omega (q-1) \geq 50$$. For $$14\leq \omega (q-1) \leq 49,$$ we verify easily that (3.3) holds with $$s=8$$. As an example, consider $$\omega (q-1)=14$$ and $$s=8$$, whence $$\delta \geq 1-3(\frac {1}{17}+ \frac {1}{19}+ \frac {1}{23}+ \frac {1}{29}+ \frac {1}{31}+ \frac {1}{37}+ \frac {1}{41}+ \frac {1}{43})>0.1109$$. It follows that the right-hand side of (3.3) is slightly larger than $$12\,039\,747\,811\,470\,119$$, which is smaller than $$P_{14}$$. It follows that $$\mathbb {F}_q$$ has three consecutive primitive elements when $$\omega (q-1)=14$$. We are unable to proceed directly when $$\omega (q-1) \leq 13$$. For example, when $$\omega (q-1) = 13$$ there is no value of $$s$$ with $$1\leq s \leq 13$$ that resolves (3.3). If we choose $$s=8,$$ then we minimize the right-hand side of (3.3) whence it follows that we need only consider $$\omega (q-1) \leq 13$$ and $$q\leq 3.49\times 10^{15}$$. We continue this procedure for larger values of $$n$$. We use Theorem 3 to obtain an initial bound on $$\omega (q-1)$$, then use Theorem 5, with suitable values of $$s$$, to reduce this bound as far as possible. We therefore reduce the problem of finding $$n$$ consecutive primitive elements in $$\mathbb {F}_{q}$$ to the finite computation in which we need only check those $$q$$ in a certain range: this range is given in Table 1. We now use Theorem 3 to obtain a bound on $$q_{0}(n)$$ for a generic value of $$n$$. To bound $$\omega (q-1)$$ we use Robin's result [8, Theorem 11] that $$\omega (n) \leq 1.38402\log n/(\log \log n)$$ for all $$n\geq 3$$. Since the function $$\log x /(\log \log x)$$ is increasing for $$x\geq e^{e}$$ we have   \[\omega (q-1) \leq \frac {1.38402\log q}{\log \log q},\] (4.1) for all $$q\geq 17$$. It is easy to check that (4.1) holds also for all $$3\leq q \leq 17$$. We use (4.1) to rearrange the condition in (2.2), showing that   \[\log q \left \{1 - \frac {2.76804 n \log 2}{\log \log q}\right \}\geq 2 \log (n-1).\] (4.2) We solve (4.2) by first insisting that the term in braces be bounded below by $$d$$, where $$d\in (0, 1)$$, and then insisting that $$d \log q \geq 2\log (n-1)$$. This shows that (2.2) is certainly true provided that   \[q \geq \max \{(n-1)^{2/d}, \exp (2^{2.76804 n/(1-d)})\}.\] (4.3) We choose $$d = 0.0001$$, so that we require $$q \geq \exp (2^{2.77 n})$$ for all $$n\geq 6$$. This proves Theorem 2. We remark that were one to use [8, Theorem 12] one could replace the bound in (4.1) by $$\log q/\log \log q +1.458 \log q /(\log \log q)^{2}$$. This would show, when $$n$$ is sufficiently large, that the exponent in Theorem 2 could be reduced from $$2.77$$ to $$2+ \epsilon$$ for any positive $$\epsilon$$: we have not pursued this. 5. Three consecutive primitive elements To prove Theorem 1, we verified numerically the existence of three consecutive primitive elements for all values of $$q$$ that remained after the application of Theorem 5. As explained in Section 4, for $$n=3$$ it is only necessary to consider the cases where $$\omega (q-1)\leq 13$$. For each possible value of $$\omega (q-1)$$ Theorem 5 was used to compute a bound on the values of $$q$$ below which the existence of three consecutive primitive elements was not ensured; these upper bounds are presented in the second column of Table 2. Algorithm 1 was then used to generate the values of $$q$$ that required testing. Table 2. Bounds and number of tests performed when $$n=3$$. $$\omega (q-1)$$  $$q$$ upper bound ($$M$$)  $$m+1$$ tests  $$m+1$$ survivors  $$p$$ tests  $$q$$ tests  $$1$$  $$256$$  $$7$$  $$7$$  $$3$$  $$1$$  $$2$$  $$16\,384$$  $$2425$$  $$805$$  $$164$$  $$8$$  $$3$$  $$802\,816$$  $$172\,827$$  $$21\,350$$  $$4785$$  $$26$$  $$4$$  $$31\,719\,424$$  $$5\,459\,954$$  $$149\,265$$  $$33\,357$$  $$106$$  $$5$$  $$368\,212\,715$$  $$30\,738\,304$$  $$695\,172$$  $$159\,618$$  $$236$$  $$6$$  $$9\,777\,432\,663$$  $$278\,578\,984$$  $$1\,680\,653$$  $$380\,984$$  $$405$$  $$7$$  $$48\,913\,046\,416$$  $$262\,182\,675$$  $$2\,131\,439$$  $$478\,146$$  $$353$$  $$8$$  $$327\,363\,505\,978$$  $$218\,209\,768$$  $$2\,162\,062$$  $$476\,569$$  $$203$$  $$9$$  $$6\,245\,429\,709\,655$$  $$479\,005\,331$$  $$897\,028$$  $$194\,276$$  $$63$$  $$10$$  $$22\,053\,999\,260\,750$$  $$68\,795\,792$$  $$262\,534$$  $$55\,943$$  $$9$$  $$11$$  $$117\,121\,857\,096\,884$$  $$9\,250\,747$$  $$93\,920$$  $$19\,315$$  $$1$$  $$12$$  $$1\,307\,042\,588\,523\,590$$  $$2\,378\,985$$  $$6566$$  $$1294$$  $$0$$  $$13$$  $$3\,489\,135\,957\,826\,319$$  $$11\,547$$  $$964$$  $$187$$  $$0$$  $$\omega (q-1)$$  $$q$$ upper bound ($$M$$)  $$m+1$$ tests  $$m+1$$ survivors  $$p$$ tests  $$q$$ tests  $$1$$  $$256$$  $$7$$  $$7$$  $$3$$  $$1$$  $$2$$  $$16\,384$$  $$2425$$  $$805$$  $$164$$  $$8$$  $$3$$  $$802\,816$$  $$172\,827$$  $$21\,350$$  $$4785$$  $$26$$  $$4$$  $$31\,719\,424$$  $$5\,459\,954$$  $$149\,265$$  $$33\,357$$  $$106$$  $$5$$  $$368\,212\,715$$  $$30\,738\,304$$  $$695\,172$$  $$159\,618$$  $$236$$  $$6$$  $$9\,777\,432\,663$$  $$278\,578\,984$$  $$1\,680\,653$$  $$380\,984$$  $$405$$  $$7$$  $$48\,913\,046\,416$$  $$262\,182\,675$$  $$2\,131\,439$$  $$478\,146$$  $$353$$  $$8$$  $$327\,363\,505\,978$$  $$218\,209\,768$$  $$2\,162\,062$$  $$476\,569$$  $$203$$  $$9$$  $$6\,245\,429\,709\,655$$  $$479\,005\,331$$  $$897\,028$$  $$194\,276$$  $$63$$  $$10$$  $$22\,053\,999\,260\,750$$  $$68\,795\,792$$  $$262\,534$$  $$55\,943$$  $$9$$  $$11$$  $$117\,121\,857\,096\,884$$  $$9\,250\,747$$  $$93\,920$$  $$19\,315$$  $$1$$  $$12$$  $$1\,307\,042\,588\,523\,590$$  $$2\,378\,985$$  $$6566$$  $$1294$$  $$0$$  $$13$$  $$3\,489\,135\,957\,826\,319$$  $$11\,547$$  $$964$$  $$187$$  $$0$$  View Large Lines $$10$$–$$12$$ of Algorithm 1 ensure that $$\gcd (m_{d-1},u_{i_d})=1$$ whenever line $$13$$ is reached, and ensure that $$\omega (m_d)=\omega (m_{d-1})+1$$ (that is, that $$\omega (m_d)=d$$) every time line $$17$$ is reached. Testing $$m+1$$, with either $$m=m_1$$ or $$m=m_d$$, amounts to verifying that $$m+1$$ is an odd prime power. If so, we see whether Theorem 5 can deal with it; if not, we verify that there exist three consecutive primitive elements in the corresponding finite field. It turned out to be faster to rule out values of $$m+1$$ using all possible values of $$s$$ in Theorem 5 (treating $$m+1$$ as if it were a prime power) before testing if it were a prime or a prime power. Algorithm 1. Enumeration of all odd integers $$m+1$$ that satisfy the conditions $$m\lt M$$ and $$\omega (m)=w$$.     View Large Algorithm 1 was coded using the PARI/GP calculator programming language [10] (version 2.7.2 using a GMP 6.0.0 kernel) and was run for $$w=1,2,\ldots ,13$$ on one core of a $$3.3$$GHz Intel i3-2120 processor. Since $$w=\omega (q-1)$$ this covers all cases that must be tested. It took about $$7$$  h to confirm that the following odd values of $$q$$ are the only odd ones for which the finite field $$\mathbb {F}_q$$ does not have three consecutive primitive elements: $$3$$, $$5$$, $$7$$, $$3^2$$, $$13$$, $$5^2$$, $$29$$, $$61$$, $$3^4$$, $$11^2$$, and $$13^2$$. For each value of $$\omega (q-1),$$ Table 2 presents the value of $$M$$ that was used (second column), the number of $$m+1$$ values that required testing (third column), the number of $$m+1$$ values that survived an application of Theorem 5 (fourth column), the number of these that were actually primes (fifth column, $$1\,804\,641$$ in total), and the number of these that were actually prime powers (sixth column, $$1411$$ in total). 6. Conjectures Based on numerical experiments up to $$10^8$$, the following conjectures appear to be plausible. Conjecture 1 The finite field $$\mathbb {F}_q$$ has four consecutive primitive elements except when $$q$$ is divisible by $$2$$ or by $$3$$, or when $$q$$ is one of the following: $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$5^2$$, $$29$$, $$31$$, $$41$$, $$43$$, $$61$$, $$67$$, $$71$$, $$73$$, $$79$$, $$113$$, $$11^2$$, $$13^2$$, $$181$$, $$199$$, $$337$$, $$19^2$$, $$397$$, $$23^2$$, $$571$$, $$1093$$, $$1381$$, $$7^4=2401$$. At present this conjecture is very difficult to settle by computation: there are simply too many cases to test. Consider, for example, $$\omega (q-1)=12$$. Even though, according to Table 1 we need to go up only to $$\omega (q-1)=23$$, the hard cases are the intermediate values of $$\omega (q-1)$$. It is necessary to test values of $$q$$ up to about $$4\times 10^{21}$$ that can have a prime power factor up to about $$2\times 10^{10}$$. Given the large sizes of these numbers, the small savings gained from just considering $$q\equiv 3 \pmod 4$$ and Theorem 6 will not be sufficient. Conjecture 2 The finite field $$\mathbb {F}_q$$ has five consecutive primitive elements except when $$q$$ is divisible by $$2$$ or by $$3$$, or when $$q$$ is one of the following: $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$5^2$$, $$29$$, $$31$$, $$37$$, $$41$$, $$43$$, $$47$$, $$7^2$$, $$61$$, $$67$$, $$71$$, $$73$$, $$79$$, $$101$$, $$109$$, $$113$$, $$11^2$$, $$5^3$$, $$127$$, $$131$$, $$139$$, $$151$$, $$157$$, $$163$$, $$13^2$$, $$181$$, $$193$$, $$199$$, $$211$$, $$229$$, $$241$$, $$271$$, $$277$$, $$281$$, $$17^2$$, $$307$$, $$313$$, $$331$$, $$337$$, $$19^2$$, $$379$$, $$397$$, $$433$$, $$439$$, $$461$$, $$463$$, $$23^2$$, $$547$$, $$571$$, $$577$$, $$601$$, $$613$$, $$5^4$$, $$631$$, $$691$$, $$751$$, $$757$$, $$29^2$$, $$31^2$$, $$1009$$, $$1021$$, $$1033$$, $$1051$$, $$1093$$, $$1201$$, $$1297$$, $$1321$$, $$1381$$, $$1453$$, $$1471$$, $$1489$$, $$1531$$, $$1597$$, $$1621$$, $$1723$$, $$1741$$, $$1831$$, $$43^2$$, $$1861$$, $$1933$$, $$2017$$, $$2161$$, $$2221$$, $$2311$$, $$2341$$, $$7^4$$, $$3061$$, $$59^2$$, $$3541$$, $$3571$$, $$61^2$$, $$4201$$, $$4561$$, $$4789$$, $$4831$$, $$71^2$$, $$5281$$, $$5881$$, $$89^2$$, $$8821$$, $$9091$$, $$9241$$, $$113^2$$, $$5^6=15\,625$$. We also examined fields that do not have six, seven and eight consecutive primitive elements. Since there are many of these we do not list them as in Conjectures 1 and 2 but merely indicate the last one we found. Conjecture 3 The finite field $$\mathbb {F}_q$$ has six consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, or by $$5$$, and when $$q>65\,521$$. Conjecture 4 The finite field $$\mathbb {F}_q$$ has seven consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, or by $$5$$, and when $$q>1\,037\,401$$. Conjecture 5 The finite field $$\mathbb {F}_q$$ has eight consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, by $$5$$, or by $$7$$, and when $$q>4\,476\,781$$. References 1 Carlitz L., ‘ Sets of primitive roots’, Compos. Math.  13 ( 1956) 65– 70. 2 Cohen S. D., ‘ Consecutive primitive roots in a finite field’, Proc. Amer. Math. Soc.  93 ( 1985) 189– 197. Google Scholar CrossRef Search ADS   3 Cohen S. D., ‘ Consecutive primitive roots in a finite field. II’, Proc. Amer. Math. Soc.  94 ( 1985) 605– 611. Google Scholar CrossRef Search ADS   4 Cohen S. D., ‘ Pairs of primitive roots’, Mathematika  32 ( 1985) 276– 285. Google Scholar CrossRef Search ADS   5 Cohen S. D. Oliveira e Silva T. Trudgian T. S., ‘ A proof of the conjecture of Cohen and Mullen on sums of primitive roots’, Math. Comp.  ( 2014), accepted for publication. See also arXiv:1402.2724 [math.NT]. 6 Lidl R. Niederreiter H., Finite fields , 2nd ed., Encyclopedia of Mathematics and its Applications 20 ( Cambridge University Press, Cambridge, 1997). 7 Peng C. Shen Y. Zhu Y. Liu C., ‘ A note on Weil's multiplicative character sum’, Finite Fields Appl.  35 ( 2014) 132– 133. Google Scholar CrossRef Search ADS   8 Robin G., ‘ Estimation de la fonction de Tchebychef $$\theta$$ sur le $$k$$-ième nombre premier et grandes valeurs de la fonction $$\omega (n)$$ nombre de diviseurs premiers de $$n$$’, Acta Arith.  42 ( 1983) 367– 389. 9 Tanti J. Thangadurai R., ‘ Distribution of residues and primitive roots’, Proc. Indian Acad. Sci. (Math. Sci.)  123 ( 2013) 203– 211. Google Scholar CrossRef Search ADS   10 The PARI Group, Bordeaux, ‘PARI/GP version 2.7.2’, 2014, http://pari.math.u-bordeaux.fr/. Google Scholar © 2015 London Mathematical Society http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Bulletin of the London Mathematical Society Oxford University Press

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Oxford University Press
Copyright
© 2015 London Mathematical Society
ISSN
0024-6093
eISSN
1469-2120
DOI
10.1112/blms/bdv018
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Abstract

Abstract For $$q$$ an odd prime power with $$q>169,$$ we prove that there are always three consecutive primitive elements in the finite field $$\mathbb {F}_{q}$$. Indeed, there are precisely eleven values of $$q \leq 169$$ for which this is false. For $$4\leq n \leq 8,$$ we present conjectures on the size of $$q_{0}(n)$$ such that $$q>q_{0}(n)$$ guarantees the existence of $$n$$ consecutive primitive elements in $$\mathbb {F}_{q}$$, provided that $$\mathbb {F}_{q}$$ has characteristic at least $$n$$. Finally, we improve the upper bound on $$q_{0}(n)$$ for all $$n\geq 3$$. 1. Introduction Let $$q$$ be a prime power and consider primitive elements in $$\mathbb {F}_{q}$$, the finite field of order $$q$$. Cohen [2–4] proved that $$\mathbb {F}_{q}$$ contains two consecutive distinct primitive elements whenever $$q>7$$. For $$n\geq 2,$$ we wish to determine $$q_{0}(n)$$ such that $$\mathbb {F}_{q}$$, assumed to have characteristic larger than or equal to $$n$$, contains $$n$$ consecutive distinct primitive elements for all $$q>q_{0}(n)$$. Carlitz [1] showed that $$q_{0}(n)$$ exists for all $$n$$. Tanti and Thangadurai [9, Theorem 1.3] showed that   \[q_{0}(n) \leq \exp (2^{5.54 n}), \quad (n\geq 2).\] (1.1) When $$n=3$$ this gives the enormous bound $$10^{43\,743}$$. The main point of this article is to apply techniques from [5] to prove the following result. Theorem 1 The finite field$$\mathbb {F}_{q}$$contains three consecutive primitive elements for all odd$$q>169$$. Indeed, the only fields$$\mathbb {F}_{q}$$$$($$with$$q$$odd$$)$$that do not contain three consecutive primitive elements are those for which$$q=3,5,7,9,13,25,29,61,81,121$$, or$$169$$.When $$n\geq 4$$ we improve the estimate for $$q_{0}(n)$$ in the following theorem. Theorem 2 The field$$\mathbb {F}_{q},$$assumed to have characteristic at least$$n,$$contains$$n$$consecutive primitive elements provided that$$q>q_{0}(n),$$where values of$$q_{0}(n)$$are given in the third column of Table1for$$4\leq n \leq 10$$and$$q_{0}(n)=\exp (2^{2.77 n})$$for$$n\geq 11$$. Table 1. Range of potential $$q$$ given a value of $$n$$. $$n$$  Bound on $$\omega (q-1)$$  Bound on $$q$$ ($$q_0(n)$$)  $$3$$  $$13$$  $$3.49\times 10^{15\phantom {12}}$$  $$4$$  $$23$$  $$3.29\times 10^{32\phantom {12}}$$  $$5$$  $$37$$  $$4.22\times 10^{61\phantom {12}}$$  $$6$$  $$59$$  $$4.61\times 10^{113\phantom {1}}$$  $$7$$  $$100$$  $$3.75\times 10^{220\phantom {1}}$$  $$8$$  $$171$$  $$2.27\times 10^{425\phantom {1}}$$  $$9$$  $$301$$  $$1.01\times 10^{836\phantom {1}}$$  $$10$$  $$533$$  $$6.69\times 10^{1638}$$  $$n$$  Bound on $$\omega (q-1)$$  Bound on $$q$$ ($$q_0(n)$$)  $$3$$  $$13$$  $$3.49\times 10^{15\phantom {12}}$$  $$4$$  $$23$$  $$3.29\times 10^{32\phantom {12}}$$  $$5$$  $$37$$  $$4.22\times 10^{61\phantom {12}}$$  $$6$$  $$59$$  $$4.61\times 10^{113\phantom {1}}$$  $$7$$  $$100$$  $$3.75\times 10^{220\phantom {1}}$$  $$8$$  $$171$$  $$2.27\times 10^{425\phantom {1}}$$  $$9$$  $$301$$  $$1.01\times 10^{836\phantom {1}}$$  $$10$$  $$533$$  $$6.69\times 10^{1638}$$  View Large We prove Theorem 2 and discuss the construction of Table 1 in Section 4. We remark that the outer exponent in the bound in Theorem 2 is half of that given in (1.1) owing entirely to the superior sieving inequality used in Theorem 3. The double exponent still gives an enormous bound on $$q_{0}(n)$$. Were one interested in bounds for specific values of $$n\geq 11,$$ one should extend Table 1 as per Section 4. In Section 5, we present an algorithm that, along with Theorem 2, proves Theorem 1. Whereas we are not able to resolve completely the values of $$q_{0}(n)$$ for $$n\geq 4$$, we present, in Section 6, some conjectures as to the size of $$q_{0}(n)$$ for $$4\leq n \leq 8$$. 2. Character sum expressions and estimates Let $$\omega (m)$$ denote the number of distinct prime factors of $$m$$ so that $$W(m)=2^{\omega (m)}$$ is the number of square-free divisors of $$m$$. Also, let $$\theta (m)=\prod _{p|m}(1-p^{-1})$$. For any integer $$m$$ define its radical $$\mathrm {Rad}(m)$$ as the product of all distinct prime factors of $$m$$. Let $$e$$ be a divisor of $$q-1$$. Call $$g \in \mathbb {F}_{q}$$$$e$$-free if $$g \neq 0$$ and $$g = h^d$$, where $$h \in \mathbb {F}_{q}$$ and $$d\,{|}\,e$$ implies $$d=1$$. The notion of $$e$$-free depends (among divisors of $$q-1$$) only on $$\mathrm {Rad}(e)$$. Moreover, in this terminology a primitive element of $$\mathbb {F}_{q}$$ is a $$(q-1)$$-free element. The definition of any multiplicative character $$\chi$$ on $$\mathbb {F}_{q}^* $$ is extended to the whole of $$\mathbb {F}_{q}$$ by setting $$\chi (0)=0$$. In fact, for any divisor $$d$$ of $$\phi (q-1)$$, there are precisely $$\phi (d)$$ characters of order (precisely) $$d$$, a typical such character being denoted by $$\chi _d$$. In particular, $$\chi _1$$, the principal character, takes the value $$1$$ at all non-zero elements of $$\mathbb {F}_{q}$$ (whereas $$\chi _1(0)=0$$). A convenient shorthand notation to be employed for any divisor $$e$$ of $$q-1$$ is   \[\int _{d\,{|}\,e} = \sum _{d\,{|}\,e}\frac {\mu (d)}{\phi (d)} \sum _{\chi _{d}},\] (2.1) where the sum over $$\chi _{d}$$ is the sum over all $$\phi (d)$$ multiplicative characters $$\chi _d$$ of $$\mathbb {F}_{q}$$ of exact order $$d$$, and where $$\mu (n)$$ is the Möbius function. Its significance is that, for any $$g \in \mathbb {F}_{q}$$,   \[\theta (e)\int _{d\,{|}\,e}\chi _d(g) = \left \{\begin {array}{@{}cl} 1 & \mbox {if $g$ is non-zero and $e$-free}, \\ 0 & \mbox {otherwise.} \end {array} \right .\] In this expression (and throughout) only characters $$\chi _d$$ with $$d$$ square-free contribute (even if $$e$$ is not square-free). The present investigation concerns the question of the existence of $$n \geq 2$$ consecutive distinct primitive elements in $$\mathbb {F}_{q}$$, that is, whether there exists $$g \in \mathbb {F}_{q}$$ such that $$\{g, g+1,\ldots , g+n-1\}$$ is a set of $$n$$ distinct primitive elements of $$\mathbb {F}_{q}$$. Define $$p$$ to be the characteristic of $$\mathbb {F}_q$$, so that $$q$$ is a power of $$p$$. Then, necessarily, $$n \leq p$$ and, by [5, Theorem 1], we can suppose $$n \geq 3$$. Assume therefore throughout that $$3 \leq n\leq p$$. In particular, $$q$$ is odd. Let $$e_1, \ldots , e_n$$ be divisors of $$q-1$$. (In practice all divisors will be even.) Define $$N(e_1, \ldots , e_n)$$ to be the number of (non-zero) $$g \in \mathbb {F}_{q}$$ such that $$g+k-1$$ is $$e_k$$-free for each $$k=1, \ldots , n$$. The first step is the standard expression for this quantity in terms of the multiplicative characters of $$\mathbb {F}_{q}^* $$. Lemma 1 Suppose$$3 \leq n\leq p$$and$$e_1, \ldots , e_n$$are divisors of$$q-1$$. Then  \[N(e_1, \ldots , e_n)= \theta (e_1)\cdots \theta (e_n) \int _{d_1|\,e_1}\cdots \int _{d_n|\,e_n}S(\chi _{d_1}, \ldots ,\chi _{d_n}),\] where  \[S(\chi _{d_1}, \ldots ,\chi _{d_n})= \sum _{g \in \mathbb {F}_{q}}\chi _{d_1}(g) \cdots \chi _{d_n}(g+n-1).\]We now provide a bound on the size of $$S(\chi _{d_1}, \ldots ,\chi _{d_n})$$. Lemma 2 Suppose$$d_1, \ldots , d_n$$are square-free divisors of$$q-1$$. Then  \[S(\chi _{d_1}, \ldots ,\chi _{d_n})=q-n \quad \mbox {if $d_1= \cdots = d_n=1$,}\]and otherwise  \[|S(\chi _{d_1}, \ldots ,\chi _{d_n})| \leq (n-1)\sqrt {q}.\] Proof We can assume not all of $$d_1, \ldots , d_n$$ have the value $$1$$. Then $$d:= \mathrm {lcm}(d_1, \ldots , d_n)$$ is also a square-free divisor of $$q-1$$ and $$d>1$$. Evidently, there are positive integers $$c_1,\ldots ,c_n$$ with $$\gcd (c_k,d_k)=1$$ such that   \[S(\chi _{d_1}, \ldots ,\chi _{d_n})= \sum _{g \in \mathbb {F}_{q}}\chi _{d}(f(g)),\] where $$f(x) =x^{c_1}(x+1)^{c_2} \cdots (x+n-1)^{c_n}$$. Since the radical of the polynomial $$f$$ has degree $$n$$ the result holds by Weil's theorem (see [6, Theorem 5.41, p. 225] and also [7]). When $$e_1= e_2=\cdots =e_n=e$$, say, we shall abbreviate $$N(e, \ldots ,e)$$ to $$N_n(e)$$. We obtain a lower bound for $$N_{n}(e)$$ in the following lemma. Lemma 3 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Then  \[N_n(e) \geq \theta (e)^n (q-(n-1)W(e)^n\sqrt {q}).\] Proof The presence of the Möbius function in the integral notation in (2.1) means that we need only concern ourselves with the square-free divisors $$(d_{1}, \ldots , d_{n})$$ of $$e$$. The contribution of each $$n$$-tuple is given by Lemma 2. Hence, $$N_n(e) \geq \theta (e)^n (q-n-(n-1)(W(e)^n-1)\sqrt {q})$$. The result will follow provided that $$\sqrt {q}(n-1) - n$$ is positive. This is easy to see, since $$n\geq 3$$ and $$q \geq 3$$. Applying Lemma 3 with $$e=q-1$$ gives the basic criterion that guarantees $$n$$ consecutive primitive elements for sufficiently large $$q$$. Theorem 3 Suppose$$3 \leq n\leq p$$. Suppose  \[q \geq (n-1)^2\,W(q-1)^{2n}=(n-1)^2\,2^{2n\omega (q-1)}.\] (2.2)Then there exists a set of$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. As an application of Theorem 3, consider the case $$n=3$$. Let $$P_m$$ be the product of the first $$m$$ primes. A quick numerical computation reveals that for $$m\geq 50$$ one has $$P_m+1 \geq 2^{2+6m}$$. Hence, it follows that $$\mathbb {F}_q$$ has three consecutive primitive elements when $$\omega (q-1)\geq 50$$ or when $$q\geq 2^{2+6\times 50}$$. We shall soon improve this markedly. We now briefly present an improvement in the above discussion when $$q \equiv 3\pmod 4$$. The improvement in this case, in which $$-1$$ is a non-square in $$\mathbb {F}_{q}$$, is related to a device used in [4]. Note that, in Lemma 1, in the definition of $$S(\chi _{d_1}, \ldots ,\chi _{d_n})$$ we may replace $$g$$ by $$-g-(n-1)$$ and obtain   \[\begin {array}{rl} S(\chi _{d_1}, \ldots ,\chi _{d_n}) & = \sum _{g \in \mathbb {F}_{q}} \prod _{i=1}^n \chi _{d_i}(-g-n+1+i-1)= \sum _{g \in \mathbb {F}_{q}} \prod _{i=1}^n \chi _{d_i}(-1)\chi _{d_i}(g+n-i) \\ & = (\chi _{d_1}\chi _{d_2}\cdots \chi _{d_n})(-1) S(\chi _{d_n}, \ldots ,\chi _{d_1}). \end {array}\] In particular, if an odd number of the integers $$d_1, \ldots , d_n$$ are even, then $$S(\chi _{d_n}, \ldots ,\chi _{d_1})=-S(\chi _{d_1}, \ldots ,\chi _{d_n})$$. As a consequence, in Lemma 1, if $$e_1= \cdots =e_n=e$$ is even, then, on the right-hand side of the expression for $$N_n(e)$$, the terms corresponding to divisors $$(d_1, \ldots , d_n)$$ with an odd number of even divisors cancel out exactly and only those with an even number of even divisors contribute. This accounts for precisely half the $$n$$-tuples $$(d_1, \ldots , d_n)$$. Accordingly, we obtain the following improvement of Lemma 3 and Theorem 3 in this situation. Theorem 4 Suppose$$3\leq n\leq p$$and$$e$$is an even divisor of$$q-1,$$where$$q\equiv 3 \pmod 4$$and$$q \geq 7$$. Then  \[N_n(e) \geq \theta (e)^n \left ( q- \frac {n-1}{2}\,W(e)^n\sqrt {q}\right ) .\]Moreover, if  \[q \geq \frac {(n-1)^2}{4}W(q-1)^{2n}= (n-1)^2 2^{2(n\omega (q-1)-1)},\]then there exists a set of$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. 3. Sieving inequalities and estimates As before, assume $$3 \leq n \leq p$$ and let $$e$$ be a divisor of $$q-1$$. Given positive integers $$m, j, k$$ with $$1 \leq j,k \leq n$$ define $$m_{jk}=m$$ if $$j=k$$ and otherwise $$m_{jk}=1$$. Lemma 4 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Let$$l$$be a prime divisor of$$q-1$$not dividing$$e$$. Then, for each$$j \in \{1, \ldots , n\},$$  \[|N(l_{j_1}e, \ldots l_{jn}e) -\theta (l)N_n(e)| \leq (1-1/l)\,\theta (e)^{n}(n-1)W(e)^n \sqrt {q}.\] Proof Observe that $$\theta (le)= (1-1/l)\,\theta (e)$$ and that, by Lemma 1, all the character sums in $$\theta (l)N_n(e)$$ appear identically in $$N(l_{j_1}e, \ldots l_{jn}e)$$. Hence, by Lemma 2,   \[|N(l_{j_1}e, \ldots l_{jn}e) -\theta (l)N_n(e)| \leq (1-1/l)\,\theta (e)^{n}(n-1)W(e)^{n-1} (W(le)-W(e)) \sqrt {q}\] and the result follows since $$W(le)=2W(e)$$. This leads to the main sieving result. Let $$e$$ be a divisor of $$q-1$$. In what follows, if $$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$ set $$s=0$$ and $$\delta =1$$. Otherwise, let $$p_1, \ldots , p_s$$, $$s \geq 1$$, be the primes dividing $$q-1$$ but not $$e$$, and set $$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. It is essential to choose $$e$$ so that $$\delta >0$$. Lemma 5 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. Then, with the above notation,  \[N_n(q-1) \geq \left ( \sum _{j=1}^n\sum _{i=1}^sN(p_{ij1}e, \ldots , p_{ijn}e) \right ) - (ns-1)N_n(e),\] (3.1)where$$p_{ijk}$$means$$(p_i)_{jk}$$. Hence  \[N_n(q-1) \geq \delta N_n(e) + \sum _{i=1}^s \left ( \sum _{j=1}^n N(p_{ij1}e, \ldots , p_{ijn}e) -\theta (p_i)N_n(e) \right ) .\] (3.2) Proof It suffices to suppose $$s \geq 1$$. The various $$N$$ terms on the right-hand side of (3.1) can be regarded as counting functions on the set of $$g \in \mathbb {F}_{q}$$ for which $$g+j-1$$ is $$e$$-free for each $$j=1, \ldots , n$$. In particular, $$N_n(e)$$ counts all such elements, whereas, for each $$i=1, \ldots , s$$, and $$1 \leq j \leq n$$, $$N(p_{ij1}e,\ldots , p_{ijn}e)$$ counts only those for which additionally $$g+j-1$$ is $$p_i$$-free. Note that $$N_n(q-1)$$ is the number of elements $$g$$ such that not only are $$g+j-1$$$$e$$-free for each $$j=1, \ldots , n$$ but additionally $$g+j-1$$ is $$p_i$$-free for each $$1\leq i\leq s,\ 1 \leq j \leq n$$. Hence we see that, for a given $$g \in \mathbb {F}_{q}$$, the right-hand side of (3.1) clocks up $$1$$ if $$g+k-1$$ is primitive for every $$1 \leq k\leq n$$, and otherwise contributes a non-positive (integral) quantity. This establishes (3.1). Since $$\theta (p_i)=1-1/p_i$$, the bound (3.2) is deduced simply by rearranging the right-hand side of (3.1). We are now able to provide a condition that, if satisfied, is sufficient to prove the existence of $$n$$ consecutive primitive elements. Theorem 5 Suppose$$3 \leq n\leq p$$and$$e$$is a divisor of$$q-1$$. If$$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$, then set$$s=0$$and$$\delta =1$$. Otherwise, let$$p_1, \ldots , p_s,$$$$s \geq 1,$$be the primes dividing$$q-1$$but not$$e$$and set$$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. Assume$$\delta >0$$. If also  \[q >\left ( (n-1)\left ( \frac {ns-1}{\delta }+2\right ) W(e)^n\right ) ^2,\] (3.3)then there exist$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. Proof Assume $$\delta >0$$. From (3.2) and Lemmas 3 and 4 (noting that for each $$j=1,\ldots , n$$ the contribution to (3.2) is the same),   \[\begin {array}{l} N_n(q-1) \geq \theta (e)^n\left ( \delta (q-(n-1)W(e)^n\sqrt {q})- n\smash {\sum _{i=1}^s} \left ( 1-\frac {1}{p_i}\right ) (n-1)W(e)^n\sqrt {q}\right ) \\ \quad \quad \quad \quad \quad = \delta \,\theta (e)^n\sqrt {q} \left ( \sqrt {q}-(n-1)W(e)^n-(n-1) \left ( \frac {ns-1}{\delta }+1\right ) W(e)^n\right ) . \end {array}\] The conclusion follows. We conclude this section with the slight improvement when $$q \equiv 3 \pmod 4$$. Now, when $$e$$ is even, the character sum expressions for the terms $$\sum _{j=1}^n N(p_{ij1}e, \ldots , p_{ijn}e) -\theta (p_i)N_n(e)$$ in (3.2) cancel unless an even number of divisors $$d_1, \ldots , d_n$$ are even. This reduces the bound in Lemma 4 by a factor of two, and gives the following improvement to Theorem 5. Theorem 6 Suppose$$q \equiv 3 \pmod 4,$$$$q \geq 7,$$$$3 \leq n\leq p$$and$$e$$is an even divisor of$$q-1$$. If$$\mathrm {Rad}(e)=\mathrm {Rad}(q-1)$$set$$s=0$$and$$\delta =1$$. Otherwise, let$$p_1, \ldots , p_s,$$$$s \geq 1,$$be the primes dividing$$q-1$$but not$$e$$and set$$\delta =1-n\sum _{i=1}^s p_i^{-1}$$. Assume$$\delta >0$$. If also  \[q >\left ( \frac {n-1}{2}\left ( \frac {ns-1}{\delta }+2\right ) W(e)^n\right ) ^2,\]then there exist$$n$$consecutive primitive elements in$$\mathbb {F}_{q}$$. Remark 1 In Theorems 5 and 6, for a given $$s$$ the best (largest) value of $$\delta$$ is obtained when the largest $$s$$ prime factors of $$q-1$$ are used to compute $$\delta$$. 4. Application of Theorems 3 and 5 for generic $$n;$$ proof of Theorem 2 As an application of Theorem 5, consider the case $$n=3$$. We showed after Theorem 3 that $$\mathbb {F}_{q}$$ contains three consecutive primitive elements for all $$q$$ satisfying $$\omega (q-1) \geq 50$$. For $$14\leq \omega (q-1) \leq 49,$$ we verify easily that (3.3) holds with $$s=8$$. As an example, consider $$\omega (q-1)=14$$ and $$s=8$$, whence $$\delta \geq 1-3(\frac {1}{17}+ \frac {1}{19}+ \frac {1}{23}+ \frac {1}{29}+ \frac {1}{31}+ \frac {1}{37}+ \frac {1}{41}+ \frac {1}{43})>0.1109$$. It follows that the right-hand side of (3.3) is slightly larger than $$12\,039\,747\,811\,470\,119$$, which is smaller than $$P_{14}$$. It follows that $$\mathbb {F}_q$$ has three consecutive primitive elements when $$\omega (q-1)=14$$. We are unable to proceed directly when $$\omega (q-1) \leq 13$$. For example, when $$\omega (q-1) = 13$$ there is no value of $$s$$ with $$1\leq s \leq 13$$ that resolves (3.3). If we choose $$s=8,$$ then we minimize the right-hand side of (3.3) whence it follows that we need only consider $$\omega (q-1) \leq 13$$ and $$q\leq 3.49\times 10^{15}$$. We continue this procedure for larger values of $$n$$. We use Theorem 3 to obtain an initial bound on $$\omega (q-1)$$, then use Theorem 5, with suitable values of $$s$$, to reduce this bound as far as possible. We therefore reduce the problem of finding $$n$$ consecutive primitive elements in $$\mathbb {F}_{q}$$ to the finite computation in which we need only check those $$q$$ in a certain range: this range is given in Table 1. We now use Theorem 3 to obtain a bound on $$q_{0}(n)$$ for a generic value of $$n$$. To bound $$\omega (q-1)$$ we use Robin's result [8, Theorem 11] that $$\omega (n) \leq 1.38402\log n/(\log \log n)$$ for all $$n\geq 3$$. Since the function $$\log x /(\log \log x)$$ is increasing for $$x\geq e^{e}$$ we have   \[\omega (q-1) \leq \frac {1.38402\log q}{\log \log q},\] (4.1) for all $$q\geq 17$$. It is easy to check that (4.1) holds also for all $$3\leq q \leq 17$$. We use (4.1) to rearrange the condition in (2.2), showing that   \[\log q \left \{1 - \frac {2.76804 n \log 2}{\log \log q}\right \}\geq 2 \log (n-1).\] (4.2) We solve (4.2) by first insisting that the term in braces be bounded below by $$d$$, where $$d\in (0, 1)$$, and then insisting that $$d \log q \geq 2\log (n-1)$$. This shows that (2.2) is certainly true provided that   \[q \geq \max \{(n-1)^{2/d}, \exp (2^{2.76804 n/(1-d)})\}.\] (4.3) We choose $$d = 0.0001$$, so that we require $$q \geq \exp (2^{2.77 n})$$ for all $$n\geq 6$$. This proves Theorem 2. We remark that were one to use [8, Theorem 12] one could replace the bound in (4.1) by $$\log q/\log \log q +1.458 \log q /(\log \log q)^{2}$$. This would show, when $$n$$ is sufficiently large, that the exponent in Theorem 2 could be reduced from $$2.77$$ to $$2+ \epsilon$$ for any positive $$\epsilon$$: we have not pursued this. 5. Three consecutive primitive elements To prove Theorem 1, we verified numerically the existence of three consecutive primitive elements for all values of $$q$$ that remained after the application of Theorem 5. As explained in Section 4, for $$n=3$$ it is only necessary to consider the cases where $$\omega (q-1)\leq 13$$. For each possible value of $$\omega (q-1)$$ Theorem 5 was used to compute a bound on the values of $$q$$ below which the existence of three consecutive primitive elements was not ensured; these upper bounds are presented in the second column of Table 2. Algorithm 1 was then used to generate the values of $$q$$ that required testing. Table 2. Bounds and number of tests performed when $$n=3$$. $$\omega (q-1)$$  $$q$$ upper bound ($$M$$)  $$m+1$$ tests  $$m+1$$ survivors  $$p$$ tests  $$q$$ tests  $$1$$  $$256$$  $$7$$  $$7$$  $$3$$  $$1$$  $$2$$  $$16\,384$$  $$2425$$  $$805$$  $$164$$  $$8$$  $$3$$  $$802\,816$$  $$172\,827$$  $$21\,350$$  $$4785$$  $$26$$  $$4$$  $$31\,719\,424$$  $$5\,459\,954$$  $$149\,265$$  $$33\,357$$  $$106$$  $$5$$  $$368\,212\,715$$  $$30\,738\,304$$  $$695\,172$$  $$159\,618$$  $$236$$  $$6$$  $$9\,777\,432\,663$$  $$278\,578\,984$$  $$1\,680\,653$$  $$380\,984$$  $$405$$  $$7$$  $$48\,913\,046\,416$$  $$262\,182\,675$$  $$2\,131\,439$$  $$478\,146$$  $$353$$  $$8$$  $$327\,363\,505\,978$$  $$218\,209\,768$$  $$2\,162\,062$$  $$476\,569$$  $$203$$  $$9$$  $$6\,245\,429\,709\,655$$  $$479\,005\,331$$  $$897\,028$$  $$194\,276$$  $$63$$  $$10$$  $$22\,053\,999\,260\,750$$  $$68\,795\,792$$  $$262\,534$$  $$55\,943$$  $$9$$  $$11$$  $$117\,121\,857\,096\,884$$  $$9\,250\,747$$  $$93\,920$$  $$19\,315$$  $$1$$  $$12$$  $$1\,307\,042\,588\,523\,590$$  $$2\,378\,985$$  $$6566$$  $$1294$$  $$0$$  $$13$$  $$3\,489\,135\,957\,826\,319$$  $$11\,547$$  $$964$$  $$187$$  $$0$$  $$\omega (q-1)$$  $$q$$ upper bound ($$M$$)  $$m+1$$ tests  $$m+1$$ survivors  $$p$$ tests  $$q$$ tests  $$1$$  $$256$$  $$7$$  $$7$$  $$3$$  $$1$$  $$2$$  $$16\,384$$  $$2425$$  $$805$$  $$164$$  $$8$$  $$3$$  $$802\,816$$  $$172\,827$$  $$21\,350$$  $$4785$$  $$26$$  $$4$$  $$31\,719\,424$$  $$5\,459\,954$$  $$149\,265$$  $$33\,357$$  $$106$$  $$5$$  $$368\,212\,715$$  $$30\,738\,304$$  $$695\,172$$  $$159\,618$$  $$236$$  $$6$$  $$9\,777\,432\,663$$  $$278\,578\,984$$  $$1\,680\,653$$  $$380\,984$$  $$405$$  $$7$$  $$48\,913\,046\,416$$  $$262\,182\,675$$  $$2\,131\,439$$  $$478\,146$$  $$353$$  $$8$$  $$327\,363\,505\,978$$  $$218\,209\,768$$  $$2\,162\,062$$  $$476\,569$$  $$203$$  $$9$$  $$6\,245\,429\,709\,655$$  $$479\,005\,331$$  $$897\,028$$  $$194\,276$$  $$63$$  $$10$$  $$22\,053\,999\,260\,750$$  $$68\,795\,792$$  $$262\,534$$  $$55\,943$$  $$9$$  $$11$$  $$117\,121\,857\,096\,884$$  $$9\,250\,747$$  $$93\,920$$  $$19\,315$$  $$1$$  $$12$$  $$1\,307\,042\,588\,523\,590$$  $$2\,378\,985$$  $$6566$$  $$1294$$  $$0$$  $$13$$  $$3\,489\,135\,957\,826\,319$$  $$11\,547$$  $$964$$  $$187$$  $$0$$  View Large Lines $$10$$–$$12$$ of Algorithm 1 ensure that $$\gcd (m_{d-1},u_{i_d})=1$$ whenever line $$13$$ is reached, and ensure that $$\omega (m_d)=\omega (m_{d-1})+1$$ (that is, that $$\omega (m_d)=d$$) every time line $$17$$ is reached. Testing $$m+1$$, with either $$m=m_1$$ or $$m=m_d$$, amounts to verifying that $$m+1$$ is an odd prime power. If so, we see whether Theorem 5 can deal with it; if not, we verify that there exist three consecutive primitive elements in the corresponding finite field. It turned out to be faster to rule out values of $$m+1$$ using all possible values of $$s$$ in Theorem 5 (treating $$m+1$$ as if it were a prime power) before testing if it were a prime or a prime power. Algorithm 1. Enumeration of all odd integers $$m+1$$ that satisfy the conditions $$m\lt M$$ and $$\omega (m)=w$$.     View Large Algorithm 1 was coded using the PARI/GP calculator programming language [10] (version 2.7.2 using a GMP 6.0.0 kernel) and was run for $$w=1,2,\ldots ,13$$ on one core of a $$3.3$$GHz Intel i3-2120 processor. Since $$w=\omega (q-1)$$ this covers all cases that must be tested. It took about $$7$$  h to confirm that the following odd values of $$q$$ are the only odd ones for which the finite field $$\mathbb {F}_q$$ does not have three consecutive primitive elements: $$3$$, $$5$$, $$7$$, $$3^2$$, $$13$$, $$5^2$$, $$29$$, $$61$$, $$3^4$$, $$11^2$$, and $$13^2$$. For each value of $$\omega (q-1),$$ Table 2 presents the value of $$M$$ that was used (second column), the number of $$m+1$$ values that required testing (third column), the number of $$m+1$$ values that survived an application of Theorem 5 (fourth column), the number of these that were actually primes (fifth column, $$1\,804\,641$$ in total), and the number of these that were actually prime powers (sixth column, $$1411$$ in total). 6. Conjectures Based on numerical experiments up to $$10^8$$, the following conjectures appear to be plausible. Conjecture 1 The finite field $$\mathbb {F}_q$$ has four consecutive primitive elements except when $$q$$ is divisible by $$2$$ or by $$3$$, or when $$q$$ is one of the following: $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$5^2$$, $$29$$, $$31$$, $$41$$, $$43$$, $$61$$, $$67$$, $$71$$, $$73$$, $$79$$, $$113$$, $$11^2$$, $$13^2$$, $$181$$, $$199$$, $$337$$, $$19^2$$, $$397$$, $$23^2$$, $$571$$, $$1093$$, $$1381$$, $$7^4=2401$$. At present this conjecture is very difficult to settle by computation: there are simply too many cases to test. Consider, for example, $$\omega (q-1)=12$$. Even though, according to Table 1 we need to go up only to $$\omega (q-1)=23$$, the hard cases are the intermediate values of $$\omega (q-1)$$. It is necessary to test values of $$q$$ up to about $$4\times 10^{21}$$ that can have a prime power factor up to about $$2\times 10^{10}$$. Given the large sizes of these numbers, the small savings gained from just considering $$q\equiv 3 \pmod 4$$ and Theorem 6 will not be sufficient. Conjecture 2 The finite field $$\mathbb {F}_q$$ has five consecutive primitive elements except when $$q$$ is divisible by $$2$$ or by $$3$$, or when $$q$$ is one of the following: $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$5^2$$, $$29$$, $$31$$, $$37$$, $$41$$, $$43$$, $$47$$, $$7^2$$, $$61$$, $$67$$, $$71$$, $$73$$, $$79$$, $$101$$, $$109$$, $$113$$, $$11^2$$, $$5^3$$, $$127$$, $$131$$, $$139$$, $$151$$, $$157$$, $$163$$, $$13^2$$, $$181$$, $$193$$, $$199$$, $$211$$, $$229$$, $$241$$, $$271$$, $$277$$, $$281$$, $$17^2$$, $$307$$, $$313$$, $$331$$, $$337$$, $$19^2$$, $$379$$, $$397$$, $$433$$, $$439$$, $$461$$, $$463$$, $$23^2$$, $$547$$, $$571$$, $$577$$, $$601$$, $$613$$, $$5^4$$, $$631$$, $$691$$, $$751$$, $$757$$, $$29^2$$, $$31^2$$, $$1009$$, $$1021$$, $$1033$$, $$1051$$, $$1093$$, $$1201$$, $$1297$$, $$1321$$, $$1381$$, $$1453$$, $$1471$$, $$1489$$, $$1531$$, $$1597$$, $$1621$$, $$1723$$, $$1741$$, $$1831$$, $$43^2$$, $$1861$$, $$1933$$, $$2017$$, $$2161$$, $$2221$$, $$2311$$, $$2341$$, $$7^4$$, $$3061$$, $$59^2$$, $$3541$$, $$3571$$, $$61^2$$, $$4201$$, $$4561$$, $$4789$$, $$4831$$, $$71^2$$, $$5281$$, $$5881$$, $$89^2$$, $$8821$$, $$9091$$, $$9241$$, $$113^2$$, $$5^6=15\,625$$. We also examined fields that do not have six, seven and eight consecutive primitive elements. Since there are many of these we do not list them as in Conjectures 1 and 2 but merely indicate the last one we found. Conjecture 3 The finite field $$\mathbb {F}_q$$ has six consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, or by $$5$$, and when $$q>65\,521$$. Conjecture 4 The finite field $$\mathbb {F}_q$$ has seven consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, or by $$5$$, and when $$q>1\,037\,401$$. Conjecture 5 The finite field $$\mathbb {F}_q$$ has eight consecutive primitive elements when $$q$$ is not divisible by $$2$$, by $$3$$, by $$5$$, or by $$7$$, and when $$q>4\,476\,781$$. References 1 Carlitz L., ‘ Sets of primitive roots’, Compos. Math.  13 ( 1956) 65– 70. 2 Cohen S. D., ‘ Consecutive primitive roots in a finite field’, Proc. Amer. Math. Soc.  93 ( 1985) 189– 197. Google Scholar CrossRef Search ADS   3 Cohen S. D., ‘ Consecutive primitive roots in a finite field. II’, Proc. Amer. Math. Soc.  94 ( 1985) 605– 611. Google Scholar CrossRef Search ADS   4 Cohen S. D., ‘ Pairs of primitive roots’, Mathematika  32 ( 1985) 276– 285. Google Scholar CrossRef Search ADS   5 Cohen S. D. Oliveira e Silva T. Trudgian T. S., ‘ A proof of the conjecture of Cohen and Mullen on sums of primitive roots’, Math. Comp.  ( 2014), accepted for publication. See also arXiv:1402.2724 [math.NT]. 6 Lidl R. Niederreiter H., Finite fields , 2nd ed., Encyclopedia of Mathematics and its Applications 20 ( Cambridge University Press, Cambridge, 1997). 7 Peng C. Shen Y. Zhu Y. Liu C., ‘ A note on Weil's multiplicative character sum’, Finite Fields Appl.  35 ( 2014) 132– 133. Google Scholar CrossRef Search ADS   8 Robin G., ‘ Estimation de la fonction de Tchebychef $$\theta$$ sur le $$k$$-ième nombre premier et grandes valeurs de la fonction $$\omega (n)$$ nombre de diviseurs premiers de $$n$$’, Acta Arith.  42 ( 1983) 367– 389. 9 Tanti J. Thangadurai R., ‘ Distribution of residues and primitive roots’, Proc. Indian Acad. Sci. (Math. Sci.)  123 ( 2013) 203– 211. Google Scholar CrossRef Search ADS   10 The PARI Group, Bordeaux, ‘PARI/GP version 2.7.2’, 2014, http://pari.math.u-bordeaux.fr/. Google Scholar © 2015 London Mathematical Society

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Published: Mar 31, 2015

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